2
$\begingroup$

Let $X$ be a metrizable space and $\{x_n\}$ be a sequence in $X$. Suppose the sequence of Dirac measures $\delta_{x_n} \xrightarrow{w} P$ where $P$ is some probability measure. Prove that $P = \delta_x$ for some $x \in X$.

I am not sure how to go about proving this exactly...

The definition of weak convergence I am using is the standard one: so $\delta_{x_n} \xrightarrow{w} P$ happens when for every continuous and bounded $f: X \xrightarrow{} \mathbb{R}$ we have $\int f d\delta_{x_n} \xrightarrow{} \int f dP$.

$\endgroup$
1
$\begingroup$

We can follow (and detail) the following steps.

  1. Fix an integer $r$ and construct a continuous and bounded function $f$ such that $f(x)=1$ if $x$ belongs to the closure of $F_r:=\bigcap_{n\geqslant r}\{x_n\}$.
  2. The definition of weak convergence implies that for each $r$, $\mathbb P(\overline{F_r})=1$.
  3. In particular, $\bigcap_r\overline{F_r}$ is non empty. It remains to prove that this set cannot contain two distinct elements
$\endgroup$
0
$\begingroup$

I saw this exercise in Billingsley's book (Convergence of Probability Measures). Here is my solution.

First, if $x_n$ has a convergent subsequence, which we identify by $x_{n_k}\to x$, then it is easy to see that $x_{n_k}\to x \Rightarrow [\delta_{x_{n_k}}\Rightarrow \delta_x]$. Since, $\delta_{x_{n_k}}\Rightarrow P$, we must have $P=\delta_x$.

Now, suppose that $x_n$ does not have any convergent subsequence. Then, it does not have a limit. In particular, none of $x_k$'s can be its limit, hence, for every given $x_k$, there is an $\epsilon_k>0$, such that for a subsequence $\{x_{n_i}\}_{i=1}^\infty$ the metric ball $B(x_k,\epsilon_k)$ does not contain any of $x_{n_i}$'s (since $d(x_k,x_{n_i})\geq\epsilon_k,\forall i$).

Since the ball $B(x_k,\epsilon_k)$ is open, we have, by Portmanteau's theorem that, $$ 0=\liminf_{n\to\infty}\delta_{x_n}(B(x_k,\epsilon_k))\geq P(B(x_k,\epsilon_k)). $$ Next, let $\displaystyle F=\bigcup_n \{x_n\}$. Observe that $F$ does not have a limit point (otherwise, we would be able to identify a subsequence converging to this limit point, contradicting with the hypothesis that $x_n$ does not have any convergent subsequence), hence $F$ must be closed. Therefore, one more application of Portmanteau's theorem gives us, $$ 1=\limsup_{n\to\infty}\delta_{x_n}(F)\leq P(F)\implies P(F)=1. $$ Finally, since $\displaystyle F\subset \bigcup_n B(x_n,\epsilon_n)$, and since, $$ 1=P(F)\leq P\left(\bigcup_n B(x_n,\epsilon_n)\right)\leq \sum_{n}P(B(x_n,\epsilon_n))=0, $$ we get a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.